Leetcode Next Permutation

最新推荐文章于 2021-03-22 19:48:08 发布

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本文介绍了一个中等难度的算法问题——实现下一个排列。该问题要求在原地修改输入数组，使其变为字典序中下一个更大的排列。若不存在更大的排列，则将其重新排列为最小的排列（即升序排列）。文章提供了详细的解决方案及示例。

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Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

Difficulty: Medium

Solution: http://www.programcreek.com/2014/06/leetcode-next-permutation-java/

public class Solution {
    public void reverse(int[] nums, int left, int right){
        while(right >= left){
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
        return;
    }
    public void nextPermutation(int[] nums) {
        int len = nums.length;
        int p = 0;
        for(int i = len - 2; i >= 0; i--){
            if(nums[i] < nums[i + 1]){
                p = i;
                break;
            }
        }
        int q = 0;
        for(int i = len - 1; i >= 0; i--){
            if(nums[i] > nums[p]){
                q = i;
                break;
            }
        }
        
        if(p == 0 && q == 0){
            reverse(nums, 0, len - 1);
            return;
        }
        int temp = nums[p];
        nums[p] = nums[q];
        nums[q] = temp;
        if(p < len)
            reverse(nums, p + 1, len - 1);
    }
}